Spring 2016, problem 8

Find all continuous functions $f : \mathbf{R}\to\mathbf{R}$ such that $$f (f (x)) = f (x)+x,$$ for any $x\in \mathbf{R}$.


5 years ago

Hi. I am Debinson from Bogotá-Colombia f(f(x))=f(x)+x

Previous ideas: We find the solution of the next equation: f(f(x))=f(x) In this case we would have that f(x) must be a constant function or identity function. We find the solution of the next equation:


In this case we would have that f(x) must be identity function. Thus the function f(x) must be of the form f(x)=ax and we would have: f(f(x))=f(x)+x a^2 x=ax+x Which bring us: x∙(a^2-a-1)=0 And we get: a=(1±√5)/2 Requested functions are: f(x)=(1±√5)/2 x

I understand your solution but I was having a problem that you did X=10Y+R, why you have taken 10 in particular, I mean you can take any other integer like 9,11 etc.

countess179 1 month ago
6 years ago

f(x) has to be a linear function, or else f(f(x)) cannot be of the same order as f(x) + x which is required by the given equation. Once we observe this, it's straightforward what this linear function is. f(x) = [(1+/-sqrt(5))/2]x.

6 years ago

Hi, I'm Hubert from Paris ...

  • $f$ is injective by $g_x=f_x-x\Rightarrow~g(f_x)=x$, and so continuously monotonic, $\lim_{+\infty} f=\alpha$ contradicts $(*)$ hence $|\lim_{+\infty}f|=\infty$, by same way $|\lim_{-\infty}f|=\infty$; hence $f$ is surjective and one to one, as $\phi=f^{-1}$,

$(*) f(f_x)=f_x+x\Rightarrow~\phi^2_x+\phi_x=x$


with $-b=\frac 1a=\varphi=\frac{\sqrt 5+1}2$, solving yields $ A\sqrt 5=bx-\phi_x; B\sqrt 5=\phi_x-ax$;

$f_0=0$ by $f_0>0\Rightarrow~0+f_0=f^2_0>f_0$: contradiction; same way for $f_0 \< 0$, $x>0\Rightarrow~x-\phi_x=\phi^2_x>0\Rightarrow~\phi_x\<x\rightarrow~>

$B\sqrt 5=\phi_x-ax=0 ~(|b|>1\Rightarrow |b^n|\longrightarrow\infty)\Rightarrow~f_x=\frac xa=\varphi.x$

which works, same way for $x\<0~$,

EDITED - Right ...this is only the increasing solution, same idea for the decreasing one ...

6 years ago

Two such functions are given by $ f(x):= ax$ where either $ a=\phi$ or $ a=- \frac{1}{\phi}$ with $\phi$ the golden ratio. But I don't think those are the only examples.

edit: I was mistaken. These are the only examples. Heres a proof, similiar, i guess, to huberts, which, I admit, I find impossible to follow (and something's missing there, anyway, because $f(x) = \phi x$ is not the only solution).

Consider the basic equation $f(f(x)) - f(x) = x$. Since the right hand side depends on $x$, but the LHS only depends on $f(x)$, $f$ has to be injective and therefore strictly monotonous. Furthermore, as hubert correctly states, $|\lim_{x \to \infty} f(x)| \lt \infty$ or $| \lim_{x \to -\infty} f(x)| \lt \infty$ yields an immediate contradiction. Thus $f$ is surjective. The inverse $f^{-1}$ therefore exists.

For natural $n$ define as usual

  • $f^n$ is f applied $n$ times and
  • $f^{-n}$ is $f^{-1}$ applied $n$ times.
  • $f^0(x) := x$

For given $x$, the (two-sided) sequence $..., f^{-2}(x), f^{-1}(x), x, f^{1}(x), f^{2}(x), ...$ is the (two-sided) Fibonacci sequence for initial values $x$ and $f(x)$. The solution is given by: $$ f^z(x) = u \phi^z + v (-\phi)^{-z}$$ with $u$ and $v$ constants.

Now pick an arbitrary nonzero $a_0 \in \mathbb{R} \setminus \left \{ 0 \right \}$. Note that $f(a_0) \neq a_0$, as the contrary contradicts the basic equation.

Define $a_z := f^z(a_0) =: u_a \phi^z + v_a (-\phi)^{-z}$.

We cannot, without loss of generality assume, that $a_1 > a_0$, but it turns out, that no new arguments are involved for the opposite case, so let's assume that, anyway.

Now we discern two cases:

Case 1 ($f$ is strictly increasing):

In this case $\forall z \in \mathbb{Z}: a_{z} \lt a_{z + 1} $. But if $v_a$ nonzero and we let $z \rightarrow - \infty$ the term $v_a (-\phi)^{-z}$ will dominate, turning that inequality around every other time. Thus $v_a = 0$ and $f(a) = \phi a$ and we're finished.

Case 2 ($f$ is strictly decreasing):

In this case $a_z \lt a_{z+1}$ if z is even and $a_z > a_{z+1}$ otherwise. But if $u_a \neq 0$ and we let $z \rightarrow + \infty$, the term $u_a \phi^z$ will dominate, so that only one of the equalities can be satisfied for large $z$. Thus $u_a = 0$ and $f(a) = -\frac{1}{\phi} a $ and we're finished.